9v AC input is not such a good choice to power a circuit that consumes 400mA at 5v.

You would have to rectify this 9v AC using a bridge rectifier, which means you'll have a peak DC voltage of 9v x 1.414 = 12.7v , minus the losses in the two diodes in the bridge rectifier that are always on (which would be about 0.7v for each diode), so you're left with about 11 volts.  However, if the 9v AC is made using a plain transformer or a basic power adapter, that means the output voltage will often be much higher than 9v when the circuit is turned off or idle (when the current output of the transformer is very low), so you may actually have more than 9v AC coming out, could be even 10-11v AC.  At the same time, there may be variations in the mains voltage which would cause variations in the output of the transformer secondary, making the voltage less than 11v DC peak, or higher than 11v DC peak.

The basic idea is that you may have a DC voltage anywhere between 10v and about 13v at the output of the rectifier, so using a 16v rated capacitor is cutting it kind of close.  It would be safer to use a 25v rated capacitor or higher.

One more thing... like I said, after the rectifier you get a peak voltage, you need capacitors to smooth it out and get it closer to a straight DC output.  The size of the capacitor can be aproximated and depends on current and the mains frequency and other things. In this case, let's say you're in Europe so you have a mains frequency of 50 Hz and the minimum voltage you'll get after the rectifier will be 10.5v (just in case you'll sometime have a mains voltage less than 220-230v) and you need at least 400mA so let's do the math for 500mA.

The minimum capacitor size to keep the DC output above a minimum would be aproximated using this formula  C =  Current in amps  /  [  2 x Mains Frequency x ( V dc peak  - V dc minimum) ]

Since you use a 5v 7805 regulator, that means you need at least 6.5v for the regulator to work, so your formula becomes  C = 0.5 A / 2 x 50Hz x (10.5-6.5)  = 0.5 / 400 = 0.00125 Farads or 1250uF minimum to keep the voltage to at least 6.5v at any time.  So it would make sense to use a 1200uF 25v or something bigger there.

You can change the 7805 regulator to something that doesn't need as much voltage about the ouput voltage (7805 needs 1.5-2v above), and then you could use a smaller input capacitor. For example, 1117 regulators need only 0.8-1v above output to work right.

Anyway, the reason I say 9v AC is not a good choice is because the peak dc voltage would be a bit high for the input of the 7805 linear regulator, if your circuit doesn't always use 400mA.For example, if your circuit uses only 200mA and you use an input capacitor after the rectifier chosen like above, it means at only 200mA the minimum dc voltage will be much higher, probably at around 8-9v DC. In that case, the linear regulator 7805 will have to dissipate the difference between input voltage and output voltage as heat, so you'll get  (8-9v in - 5v out ) x 0.2A = 3-4v x 0.2a = 0.6-0.8 watts, which means the linear regulator will be hot.  Above around 1.2-1.5w, especially in an enclosed case with minimal ventilation, you'll need to put a heatsink on that regulator.

Unless you have something that needs AC voltage, it makes more sense to use a 7.5v - 9v DC power adapter, at least you'll know the dc voltage will always be in that range no matter the current consumption of your circuit.